## Chemistry: Molecular Approach (4th Edition)

The molar mass of titanium is 47.90 g/mol . Thus, the mass of titanium containing$2.55\,\times \,{{10}^{24}}\,atoms$ is calculated as: \begin{align} & 1\,\,mol\,\,\,Ti\,\,=\,\,47.90\,\,g \\ & 1\,\,mol\,\,Ti\,=\,\,6.022\,\times \,{{10}^{23}}\ \ Ti\,\,atoms \\ & Thus,\,\,\,6.022\,\times \,{{10}^{23}}\ \ Ti\,\,atoms\,\,=\,\,47.90\,\,g \\ & \,\,\Rightarrow 2.55\,\times \,{{10}^{24}}\,\,\,Ti\,\,atoms\,\,=\,47.90\,g\,\,\times \,\frac{2.55\,\times \,{{10}^{24}}\,\,Ti\,\,atoms}{\,6.022\,\times \,{{10}^{23}}\ \ Ti\,\,atoms\,\,} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\,\,202.8\,\,g \\ \end{align} Calculate the volume of titanium as: \begin{align} & \,\,\,\,density\,\,=\,\,\frac{mass}{volume} \\ & \Rightarrow volume\,\,=\,\frac{mass}{density} \\ \end{align} Substitute density of titanium $=4.50\ g/c{{m}^{3}}$ and mass of titanium = 202.8 g in the above expression as: $volume\,\,=\,\,\frac{202.8\,\,g}{4.50\,\,g/c{{m}^{3}}}\,\,=\,45.0\ c{{m}^{3}}$ As a titanium atom is cubical, the volume of a titanium atom is given by: \begin{align} & volume\,\,=\,\,{{\left( a \right)}^{3}}\,\,\,\,\,=45\,c{{m}^{3}} \\ & a\,\,\,\,\,\,\,\,\,\,\,=\,\,\sqrt[3]{45\,c{{m}^{3}}} \\ & a\,\,\,\,=\,\,3.556\,\,cm\,\,\approx \,3.56\,cm \\ \end{align} The edge length of a titanium cube containing $2.55\,\times \,{{10}^{24}}\,atoms$ is 3.56 cm.